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Solution of a Spring Problem by Kinetic Energy Theorem
DOI: 10.12677/OJNS.2021.96100, PDF, HTML, XML, 下载: 487  浏览: 1,016

Abstract: Spring is one of the important models in college physics teaching and learning. In this paper, a typical example of spring problems is resolved by using the kinetic energy theorem. Compared with the spring vibrator model, the idea of solving the problem is described in detail by using the theorem of mass center motion. At the same time, it is pointed out that the problem-solving idea in this paper is a further generalization of the spring vibrator model. This has certain practical significance for broadening the thinking and understanding of the kinetic energy theorem and the problems related to the spring model.

1. 引言

2. 理论模型和研究方法

Figure 1. Schematic diagram of spring model of connecting blocks at both ends.

${v}_{{m}_{1}}=\sqrt{\frac{k{m}_{2}{l}^{2}}{{m}_{1}\left({m}_{1}+{m}_{2}\right)}}$ (1)

${v}_{{m}_{2}}=\sqrt{\frac{k{m}_{1}{l}^{2}}{{m}_{2}\left({m}_{1}+{m}_{2}\right)}}$ (2)

$|{v}_{{m}_{1}}-{v}_{{m}_{2}}|=\frac{{m}_{1}+{m}_{2}}{{m}_{2}}=\sqrt{\frac{\left({m}_{1}+{m}_{2}\right)k{l}^{2}}{{m}_{1}{m}_{2}}}$ (3)

$\frac{1}{2}k{l}^{2}=\frac{1}{2}\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}{v}_{12}^{2}$ (4)

$W={\int }_{{r}_{1}}^{{r}_{2}}F\cdot \text{d}r=\frac{1}{2}m{v}_{2}^{2}-\frac{1}{2}m{v}_{1}^{2}$ (5)

${v}_{1}$${v}_{2}$ 分别对应着物体的始末速度。

${F}_{1}=-{F}_{2}⇒{m}_{1}{a}_{1}=-{m}_{2}{a}_{2}$ (6)

${m}_{1}{x}_{1}=-{m}_{2}{x}_{2}⇒|{x}_{2}|=|\frac{{m}_{1}{x}_{1}}{{m}_{2}}|$ (7)

${F}_{1}=k{x}_{1}i$ (8)

${F}_{2}={k}_{2}{x}_{1}i+{k}_{2}{x}_{2}i={k}_{2}xi\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(x={x}_{1}+{x}_{2}\right)$ (9)

$|{F}_{1}|=|{F}_{2}|$ (10)

${k}_{2}=\frac{k{m}_{2}}{{m}_{1}+{m}_{2}}$ (11)

${\int }_{{x}_{1}}^{{x}_{2}}{F}_{2}\cdot \text{d}x={\int }_{{x}_{1}}^{{x}_{2}}\frac{k{m}_{2}x}{{m}_{1}+{m}_{2}}\text{d}x=\frac{1}{2}{m}_{1}{v}_{2}^{2}-\frac{1}{2}{m}_{1}{v}_{1}^{2}$ (12)

${x}_{1}=0,{v}_{1}=0$ 时，可得：

${v}_{{m}_{1}}=\sqrt{\frac{k{m}_{2}{x}^{2}}{{m}_{1}\left({m}_{1}+{m}_{2}\right)}}$ (13)

${v}_{{m}_{2}}=\sqrt{\frac{k{m}_{1}{x}^{2}}{{m}_{2}\left({m}_{1}+{m}_{2}\right)}}$ (14)

$|{v}_{{m}_{1}}-{v}_{{m}_{2}}|=\sqrt{\frac{\left({m}_{1}+{m}_{2}\right)k{x}^{2}}{{m}_{1}{m}_{2}}}$ (15)

${F}_{2}=\frac{k{m}_{2}x}{{m}_{1}+{m}_{2}}={k}_{2}x$ (16)

$F=-kx$ (17)

3. 结论

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